Q 12-12-093JEE MainJEE Main 2021 (27 Jul, Shift 2)Medium
An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3$ eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000$ Å. What is the maximum kinetic energy of the emitted photoelectron?
Answer: (B) $1.41$ eV
Second excited state: $n = 3$, $E_3 = -\dfrac{13.6}{9} = -1.51$ eV.
Photon energy $= 3 - (-1.51) = 4.51$ eV.
Work function $\phi = \dfrac{12400}{4000} = 3.1$ eV.
$K_{\max} = 4.51 - 3.1 = 1.41$ eV.
Solution by Sreeraj P, M.Sc Physics